急需数列证明设等差数列Sm=n,Sn=m,证明Sn+m=-(m+n).

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急需数列证明设等差数列Sm=n,Sn=m,证明Sn+m=-(m+n).急需数列证明设等差数列Sm=n,Sn=m,证明Sn+m=-(m+n).急需数列证明设等差数列Sm=n,Sn=m,证明Sn+m=-(

急需数列证明设等差数列Sm=n,Sn=m,证明Sn+m=-(m+n).
急需数列证明
设等差数列Sm=n,Sn=m,证明Sn+m=-(m+n).

急需数列证明设等差数列Sm=n,Sn=m,证明Sn+m=-(m+n).
Sn=(A1+An)*n/2=m; 进而 (A1+A1+(n-1)d)*n/2=m
Sm=(A1+Am)*m/2=n; 进而 (A1+A1+(m-1)d)*m/2=n
由右边两个式子,把A1和d看做未知数,可得
2A1+(n-1)d=2m/n
2A1+(m-1)d=2n/m
令n>m,(反过来也一样),则上式减下式得
d=-2(m+n)/nm;
另一方面
Sm+n=(A1+Am+n)*(m+n)/2
=(A1+Am+n)*m/2+(A1+Am+n)*n/2
=(A1+Am+nd)*m/2+(A1+An+md)*n/2
=n+mnd/2+m+mnd/2
=m+n+nmd
将得到的结果d带入
Sn+m=m+n+(-2(m+n))=-(m+n)